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Solve the equation $$\cos x -2\cos 2x+3 \cos 3x -4\cos 4x = \dfrac{1}{2}.$$ I tried, put $t = \cos x$.

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Yes, and what happened when you did that? –  Gerry Myerson Nov 4 '12 at 3:08
    
Did you try the angle sum identity to break up some of those cosines? Remember: $cos(2x)=cos^2(x)-sin^2(x)$. –  Todd Wilcox Nov 4 '12 at 5:08
    
Or better still $\cos 2x = 2 \cos^2 x - 1$. It should be possible to turn this into a big polynomial in $\cos x$. –  user22805 Nov 4 '12 at 9:12
    
Note that $$ \begin{align} \sum\limits_{k=1}^n (-1)^k\sin kx&=\frac{1}{2}\sec\frac{x}{2}\sum\limits_{k=1}^n (-1)^k 2\sin kx\cos\frac{x}{2}\\ &=\frac{1}{2}\sec\frac{x}{2}\sum\limits_{k=1}^n (-1)^k \left(\sin\left(k+\frac{1}{2}\right)x+\sin\left(k-\frac{1}{2}\right)x\right)\\ &=\frac{1}{2}\sec\frac{x}{2}\left(-\sin\frac{x}{2}+(-1)^n\sin\left(n+\frac{1}{2}‌​\right)x\right)\\ \end{align} $$ –  userNaN Nov 4 '12 at 12:37
    
hence $$ \begin{align} \sum\limits_{k=1}^n(-1)^{k-1}k\cos kx &=-\frac{d}{dx}\left(\sum\limits_{k=1}^n (-1)^k\sin kx\right)\\ &=-\frac{d}{dx}\left(\frac{1}{2}\sec\frac{x}{2}\left(-\sin\frac{x}{2}+(-1)^n\sin‌​\left(n+\frac{1}{2}\right)x\right)\right)\\ &=\frac{1}{4}\sec^2\frac{x}{2}(1-(n+1)(-1)^n\cos nx-n(-1)^n\cos(n+1)x) \end{align} $$ Setting $n=4$ we get $$ \cos x-2\cos 2x+3\cos 3x-4\cos 4x=\frac{1}{4}\sec^2\frac{x}{2}(1-5\cos 4x-4\cos 5x) $$ –  userNaN Nov 4 '12 at 12:38

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up vote 3 down vote accepted

With $c=\cos x$ we have $\cos 2x=2c^2-1$, and $\cos 3x = 4c^3-3c$, and finally $\cos 4x = 8c^4-8c^2+1$. If you take your equation, move the 1/2 to the left side, make the above substitutions, and multiply by $-2$, you'll get $$64c^4-24c^3-56c^2+16c+5=0.$$ This factors as $(2c-1)(32c^3+4c^2-26c-5)=0$. The cubic here has no rational roots, and it looks like one would have to resort to the cubic equation (a mess) to find its zeros. Numerically the zeros of the cubic are about $-.859,-.195,+.929$, all can be cosine of an angle. So from the other root $c=1/2$ of the linear factor, you'll have a total of eight solutions in each interval $[2k \pi,(2k+2)\pi]$. Looks like only the solutions from $ \cos x=1/2$ will be familiar angles: $\pi/3$ and $5\pi/3$ and their translates by $2 \pi n$.

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Thank you very much. –  minthao_2011 Nov 4 '12 at 15:35

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